There are three salts that contain complex ions of chromium and have the molecular formula CrCl3 6H2O. Treating 0.248g of the first salt with a strong dehydrating agent resulted in a mass loss of 0.036g. Treating 248mg of the second salt with the same dehydrating agent resulted in a mass loss of 18 mg. The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to 100.0 mL portions of 0.10 M solutions of each salt resulted in the formation of different masses of silver chloride: one solution yielded 1430 mg AgCl; another, 2860 mg AgCl; the third, 4290 mg AgCl. Two of the salts are green and one is violet. What are the probable structural formulas for these salts?
A big hint: The loss in water from the dehydration experiments will tell you which H2O molecules were "more or less free" (; i.e., in or out of the coordination sphere of the central metal). The same with the ppt of AgCl will tell you which had Cl in the coordination sphere(were covalent and would not ppt AgCl) and which were outside the sphere (and would ppt AgCl). Post your work if you get stuck.
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To determine the probable structural formulas for the salts, let's analyze the information provided step by step:
1. Start by finding the molecular formula of the hydrated salts. The molecular formula given is CrCl3 · 6H2O, which means each salt contains one chromium ion (Cr3+) and three chloride ions (Cl-) for every one molecule of the salt, along with six water molecules.
2. Determine which salt corresponds to each set of given data by comparing the information provided:
- The first salt lost 0.036g upon treatment with a dehydrating agent. This mass loss indicates the removal of water molecules. Hence, this salt is the one that lost mass, which corresponds to CrCl3 · 6H2O with a mass loss of 0.036g.
- The second salt also lost mass with a mass loss of 18 mg. This corresponds to CrCl3 · 6H2O with a mass loss of 18 mg.
- The third salt did not lose any mass when treated with the dehydrating agent. Thus, the third salt must be anhydrous (without water molecules) and corresponds to CrCl3.
3. Now we need to determine which salt corresponds to each mass of silver chloride formed upon addition of aqueous silver nitrate:
- The solution that yielded 1430 mg of AgCl corresponds to the first salt, CrCl3 · 6H2O.
- The solution that yielded 2860 mg of AgCl corresponds to the second salt, CrCl3 · 6H2O.
- The solution that yielded 4290 mg of AgCl corresponds to the anhydrous third salt, CrCl3.
4. Finally, we need to identify the colors of the salts:
- The first two salts, which correspond to CrCl3 · 6H2O, are green in color.
- The anhydrous third salt, CrCl3, is violet in color.
Therefore, using all the given information, we can conclude the probable structural formulas and colors of the salts as follows:
- The first salt: CrCl3 · 6H2O (green)
- The second salt: CrCl3 · 6H2O (green)
- The third salt: CrCl3 (violet)
To determine the probable structural formulas for the salts, let's analyze the given information step by step:
1. Mass Loss upon Dehydration:
- The first salt lost 0.036g upon dehydration.
- The second salt lost 18mg upon dehydration.
- The third salt did not lose any mass.
From this information, we can deduce that the first and second salts contain water molecules (H2O) in their structures since they experienced mass loss upon dehydration. The third salt, however, does not contain any water molecules.
2. Formation of Silver Chloride:
- Adding excess aqueous silver nitrate to each salt resulted in the formation of different masses of silver chloride (AgCl).
- One solution yielded 1430 mg AgCl.
- Another solution yielded 2860 mg AgCl.
- The third solution yielded 4290 mg AgCl.
Considering that AgCl is a precipitate formed by the reaction between chloride ions (Cl-) and silver ions (Ag+), we can conclude that the different masses of AgCl formed are directly proportional to the amount of chloride ions present in each solution. Therefore, the salt that yielded 4290 mg of AgCl contains the highest amount of chloride ions, followed by the one with 2860 mg and then 1430 mg of AgCl.
3. Colors of the Salts:
- Two of the salts are green, and one is violet.
Based on these clues, we can draw the following conclusions:
- The third salt does not contain water (anhydrous) and did not lose any mass upon dehydration. This suggests that the third salt's molecular formula is CrCl3.
- The first and second salts are complex salts containing water molecules in their structure. Let's analyze them further to determine their formulas.
For the first salt:
- Mass loss upon dehydration: 0.036g
- The formula weight of water (H2O) is approximately 18 g/mol
- Using this information, we can calculate the number of water molecules lost: 0.036g / 18g/mol = 0.002 mol of water
- Since there are six water molecules per formula unit, the total number of formula units of the first salt = 0.002 mol / 6 = 0.00033 mol
Taking into account that the molar concentration (M) is given as 0.10 M in a 100.0 mL solution, we can determine the number of moles in the solution:
- Number of moles = Molarity (M) x Volume (L)
- Number of moles = 0.10 mol/L x 0.100 L = 0.010 mol
Comparing the number of moles of the salt to the number of moles in the solution, we can find the following ratio:
- 0.00033 mol (first salt) / 0.010 mol (total moles in solution) = 0.033
Therefore, the first salt's formula is CrCl3 · 6H2O.
For the second salt:
- Mass loss upon dehydration: 18 mg
- Considering the molar mass of water (H2O) is approximately 18 g/mol, the number of moles lost = 0.018 g / 18 g/mol = 0.001 mol
- Since there are six water molecules per formula unit, the total number of formula units of the second salt = 0.001 mol / 6 = 0.0001667 mol
Using the same method as before to find the ratio:
- 0.0001667 mol (second salt) / 0.010 mol (total moles in solution) = 0.0167
Therefore, the second salt's formula is CrCl3 · 6H2O.
To summarize:
- The first salt's formula: CrCl3 · 6H2O
- The second salt's formula: CrCl3 · 6H2O
- The third salt's formula: CrCl3
Based on the color descriptions, we can deduce that the first and second salts (CrCl3 · 6H2O) are green, while the third salt (CrCl3) is violet.